To find the derivative of \( f(x) = \frac{1}{x-1} \) using **first principles**, we use the definition of the derivative:

\[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\]

For \( f(x) = \frac{1}{x-1} \), let's substitute and simplify step by step.

### Step 1: Apply the difference quotient formula

\[
f'(x) = \lim_{h \to 0} \frac{\frac{1}{(x+h)-1} - \frac{1}{x-1}}{h}
\]

This simplifies to:

\[
f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h-1} - \frac{1}{x-1}}{h}
\]

### Step 2: Simplify the difference between the two fractions

To combine the two fractions in the numerator, find a common denominator:

\[
f'(x) = \lim_{h \to 0} \frac{\frac{(x-1) - (x+h-1)}{(x-1)(x+h-1)}}{h}
\]

Simplifying the numerator:

\[
f'(x) = \lim_{h \to 0} \frac{\frac{(x-1) - (x+h-1)}{(x-1)(x+h-1)}}{h} = \lim_{h \to 0} \frac{\frac{-h}{(x-1)(x+h-1)}}{h}
\]

### Step 3: Simplify the expression

Now, cancel the \( h \) in the numerator and denominator:

\[
f'(x) = \lim_{h \to 0} \frac{-1}{(x-1)(x+h-1)}
\]

### Step 4: Take the limit as \( h \to 0 \)

Now, substitute \( h = 0 \) into the remaining expression:

\[
f'(x) = \frac{-1}{(x-1)^2}
\]

### Final Answer:
Thus, the derivative of \( f(x) = \frac{1}{x-1} \) using first principles is:

\[
f'(x) = \frac{-1}{(x-1)^2}
\]

Would you like further clarification or more details?

### Relative Questions:
1. How does the derivative of a rational function typically behave?
2. What is the general process of finding a derivative using first principles?
3. Can you explain why the common denominator is necessary when simplifying the difference of two fractions?
4. How does the first principles method compare to using derivative rules like the quotient rule?
5. What happens when \( x = 1 \) in the function \( f(x) = \frac{1}{x-1} \)?

### Tip:
When dealing with rational functions, always check for undefined points (like when the denominator is zero) before applying differentiation rules.

Learn how to derive the derivative of f(x)=1/(x-1) using first principles with a step-by-step breakdown. This calculus problem covers core concepts like limits and derivatives, suitable for students in Grades 11-12 or college.

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